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Math Help - Series Question

  1. #1
    Member RedBarchetta's Avatar
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    Series Question

    Does this series converge or diverge?

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{5\ln n}}<br />
{{\sqrt[{}]{{n^3 }}}}} <br />

    Solution

    I found that:

    <br />
\frac{1}<br />
{n} < \frac{{5\ln n}}<br />
{{\sqrt[{}]{{n^3 }}}}<br />

    Therefore the series is divergent by comparison with harmonic series?

    Thank you.
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  2. #2
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    Quote Originally Posted by RedBarchetta View Post
    Does this series converge or diverge?

    <br />
\sum\limits_{n = 1}^\infty  {\frac{{5\ln n}}<br />
{{\sqrt[{}]{{n^3 }}}}} <br />

    Solution

    I found that:

    <br />
\frac{1}<br />
{n} < \frac{{5\ln n}}<br />
{{\sqrt[{}]{{n^3 }}}}<br />

    Therefore the series is divergent by comparison with harmonic series?

    Thank you.

    Use the ratio test

    Regards,

    David
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  3. #3
    Moo
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    A Cute Angle Moo's Avatar
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    Hello,

    It is supposed to converge.

    It's a Bertrand series : Bertrand Series

    Also note that \forall \alpha > 0 ~,~ \exists N \in \mathbb{N} ~,~ \forall n > N ~,~ \ln(n)<n^\alpha

    Make \alpha=\frac 14
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by David24 View Post
    Use the ratio test

    Regards,

    David
    This is inconclusive
    Quote Originally Posted by Moo View Post
    Hello,

    It is supposed to converge.

    It's a Bertrand series : Bertrand Series

    Also note that \forall \alpha > 0 ~,~ \exists N \in \mathbb{N} ~,~ \forall n > N ~,~ \ln(n)<n^\alpha

    Make \alpha=\frac 14
    Yes Moo is correct. This is in my opinion the best method to do. To prove that the inequality that Moo alluded to is correct simply prove that the minimum value for f(n)=n^{\frac{1}{4}}-\ln(n) is positive.
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  5. #5
    Member RedBarchetta's Avatar
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    This makes no sense whatsoever. The comparison test holds that I did. How in the heck can it converge?
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    This makes no sense whatsoever. The comparison test holds that I did. How in the heck can it converge?
    let f(n)=\frac{5\ln(n)}{n^{\frac{3}{2}}}-\frac{1}{n}

    You can find that f'(n)=0 at n\approx{3041}. Using the second derivative test shows that this is a minimum value. Now test this in your calculator and you will see that f\left(n>3041\right)<0\implies\forall{n}>3042~\fra  c{1}{n}\not<\frac{5\ln(n)}{n^{\frac{3}{2}}}
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  7. #7
    Member RedBarchetta's Avatar
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    Alright. That makes sense. How do you avoid the pitfall of what I did?

    How do you come up with f(n)? Take the difference between the series you're comparing?

    Thanks Mathstud for all of your posts. I'm amazed you have such a clear understanding of the material at a young age. I hope to gain this at some point, although it doesn't help when you use a crappy calculus text.
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Alright. That makes sense. How do you avoid the pitfall of what I did?
    It is usually easier if the series are algebraic, for example \frac{n+1}{n^3+3n+1} since \frac{n+1}{n^3+3n+1}\sim\frac{1}{n^2} then \exists{N}\backepsilon\forall{n\geqslant{N}}\frac{  n+1}{n^3+2n+1}\leqslant\frac{1}{an^2}

    Where a is some arbitrary constanst greater than one.

    In this case it takes a little more intuitiveness, you have to base the fact that \ln(n) is a lower order infinity than any function of the form n^{\varphi>0}, or you can write that \ln(n)\prec{n^{\varphi>0}}\implies\lim_{n\to\infty  }\frac{\ln(n)}{n^{\varphi>0}}=0. A good way of thinking about it is that n^{\varphi>0} increases quicker than \ln(n) (take the derivatives and convince yourself), so eventually even if \ln(n) may seem to be bigger, because of n^{\varphi>0}'s superior derivative it will eventually "catch up" to \ln(n) and eventually pass it. Remember \infty number of n's gives it quite a lot of time to do so.


    How do you come up with f(n)? Take the difference between the series you're comparing?
    Yes. If f(n)<g(n)~\forall{n}\in{E\subset\mathbb{R}}

    Then that means that \forall{n}\in{E\subset\mathbb{R}}~h(n)>0 where h(n)=g(n)-f(n)

    Thanks Mathstud for all of your posts. I'm amazed you have such a clear understanding of the material at a young age. I hope to gain this at some point, although it doesn't help when you use a crappy calculus text.
    Thank you
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  9. #9
    MHF Contributor Mathstud28's Avatar
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    Here is how I would formally prove this series is convergent.

    Consider the function f(n)=\sqrt[4]{n}-\ln(n), differentiation yields f'(n)=\frac{1}{4n^{\frac{3}{4}}}-\frac{1}{n}=\frac{\sqrt[4]{n}-4}{4n}. Solving this gives f(n)=0 when n=256. Now f''(256)>0 which implies that n=256 is a minimum of f(n). Now since f(256)>0\implies\forall{n}\in\mathbb{R^+}~\ln(n)\l  eqslant\sqrt[4]{n}.

    So now from that we can see that

    \forall{n}\in\mathbb{N}~0\leqslant\frac{5\ln(n)}{n  ^{\frac{3}{2}}}\leqslant\frac{5\sqrt[4]{n}}{n^{\frac{3}{2}}}=\frac{5}{n^{\frac{5}{4}}}

    So it follows that

    0\leqslant\sum_{n=2}^{\infty}\frac{5\ln(n)}{n^{\fr  ac{3}{2}}}\leqslant\sum_{n=2}^{\infty}\frac{5}{n^{  \frac{5}{4}}}

    And since \forall{n}\in\mathbb{N}~\frac{1}{n^{\frac{5}{4}}}>  0\wedge\frac{5}{n^{\frac{5}{4}}}\in\mathcal{C}\wed  ge\frac{1}{n^{\frac{5}{4}}}\in\downarrow

    So by the integral test \int_2^{\infty}\frac{5}{n^{\frac{5}{4}}}dn=10\cdot  {2^{\frac{3}{4}}} and \sum_{n=2}^{\infty}\frac{5}{n^{\frac{5}{4}}} share convergence

    Thus \sum_{n=2}^{\infty}\frac{5\ln(n)}{n^{\frac{3}{2}}} converges

    So you can model it off of that if you ever need to use it.

    NOTE: A lot of people criticize me for being too verbose, you might want to cut out some uneeded words.

    NOTE ALSO: This could have more easily been done by the integral test

    \int_2^{\infty}\frac{5\ln(n)}{n^{\frac{3}{2}}}=5\s  qrt{2}(\ln(2)+2)
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